Introducing Scooby Snacks!

rikcando, Scooby was actually correct (don't forget to mark the calendar!) about the 156 for 13 cards but only if he was using permutations, where order matters. Since order doesn't matter, combinations would be used, which brings it down to 78.

That being said, you are also correct when saying there are 52 cards, and JJ and 66 need taken into consideration - and only using 13 cards would not do that. So we'd have to calculate combinations based on 52 cards, not 13.

My previous post of:
132600 3-number combinations
2652 2-number combinations
Those are wrong as I accidentally permut'ed instead of combin'ed. Those were based on 52 cards.

So:
22100 3-number combinations
1326 2-number combinations
-----
23426 possible combinations
- ? duplicates
=====

As rikcando said, some could be doubles, so those need eliminated. For any given combination 3 of the 8 available cannot be used (the other 3 duplicates of the first number choice).

If we just calculate the actual number of winning combinations, it will automatically eliminate duplicates as being good combos. With anything as the 2nd number good, including either of the choice numbers..

Jack(1) Ace(1) 6(1)
Jack(1) Ace(2) 6(1)
etc
You get 46 for Jack(1) (any good card) 6(1)
then go on to Jack(1) Ace(1) 6(2)
Comes out to 736.

There are 16 J6 2-number combos.

Feel free to correct me

Starting to wonder, though, if we really should use permutations. Combination J-Ace-6 would not be valid as a 3-number winner if it was Ace-J-6 (it would qualify as a 2-number winner but our calculations would be all off. Not like they aren't now )

I'll check back later, gotta git bizy.

Permutations.. there are 132600 3-card permutations and 2652 2-card permutations.

4/52 of those permutations will start with J - 4/52 of *those* will end in 6
4/52 of those permutations will start with 6 - 4/52 of *those* will end in J
= 1569 3-card winning combos

16 2-card winning combos.

Too low of an expected win %, something's not quite right. I wish some of our mathematicians would chime in.

I'd like to see how you came up with 46.
52 minus all J's and 6's = 44
52 minus just this J and 6 = 50
52 minus this J and 6 and remaining 6's = 47

What did I miss?

Taking a minute out from making dinner.

Is this correct?

You can have 52 positions that a J6 could occur. Another 52 position that 6J could occur. (104)
You can have 52 positions that a Jx6 could occur. Another 52 position that 6xJ could occur. (104)
104+104=208 possible positions for a match.

Now I have 4 jack that can match with any of the four 6's giving me 16 winning card combinations.

Therefore I can have a total of 16 matches in 208 positions.
16*208 = 3328 possible winning combinations in a deck.

Now, how many total combinations in a deck?

That's what I keep coming back to as well.

So, I have 3328 chances to find 156 possible combinations. Again, I think.

So, 21.3 to 1 that I will find your cards? 95.5% seems a little high, but a little closer to what I have experienced.